Inductance, Capacitance, and Mutual Inductancefaculty.weber.edu/snaik/ECE1270/Ch6.pdfInductance,...
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6Inductance, Capacitance, and
Mutual Inductance
Assessment Problems
AP 6.1 [a] ig = 8e−300t − 8e−1200tA
v = Ldigdt
= −9.6e−300t + 38.4e−1200tV, t > 0+
v(0+) = −9.6 + 38.4 = 28.8V
[b] v = 0 when 38.4e−1200t = 9.6e−300t or t = (ln 4)/900 = 1.54ms
[c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W
[d]dp
dt= 0 when e1800t − 12.5e900t + 16 = 0
Let x = e900t and solve the quadratic x2 − 12.5x + 16 = 0
x = 1.44766, t =ln 1.45
900= 411.05µs
x = 11.0523, t =ln 11.05
900= 2.67ms
p is maximum at t = 411.05µs
[e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72W
[f ] W is max when i is max, i is max when di/dt is zero.
When di/dt = 0, v = 0, therefore t = 1.54ms.
[g] imax = 8[e−0.3(1.54) − e−1.2(1.54)] = 3.78A
wmax = (1/2)(4 × 10−3)(3.78)2 = 28.6mJ
6–1
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6–2 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
AP 6.2 [a] i = Cdv
dt= 24 × 10−6 d
dt[e−15,000t sin 30,000t]
= [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A, i(0+) = 0.72A
[b] i(
π
80ms)
= −31.66mA, v(
π
80ms)
= 20.505V,
p = vi = −649.23mW
[c] w =(
1
2
)
Cv2 = 126.13µJ
AP 6.3 [a] v =(
1
C
)∫ t
0−i dx + v(0−)
=1
0.6 × 10−6
∫ t
0−3 cos 50,000x dx = 100 sin 50,000t V
[b] p(t) = vi = [300 cos 50,000t] sin 50,000t
= 150 sin 100,000t W, p(max) = 150W
[c] w(max) =(
1
2
)
Cv2max = 0.30(100)2 = 3000µJ = 3mJ
AP 6.4 [a] Leq =60(240)
300= 48mH
[b] i(0+) = 3 + −5 = −2A
[c] i =125
6
∫ t
0+
(−0.03e−5x) dx − 2 = 0.125e−5t − 2.125A
[d] i1 =50
3
∫ t
0+
(−0.03e−5x) dx + 3 = 0.1e−5t + 2.9A
i2 =25
6
∫ t
0+
(−0.03e−5x) dx − 5 = 0.025e−5t − 5.025A
i1 + i2 = i
AP 6.5 v1 = 0.5 × 106∫ t
0+
240 × 10−6e−10x dx − 10 = −12e−10t + 2V
v2 = 0.125 × 106∫ t
0+
240 × 10−6e−10x dx − 5 = −3e−10t − 2V
v1(∞) = 2V, v2(∞) = −2V
W =[
1
2(2)(4) +
1
2(8)(4)
]
× 10−6 = 20µJ
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Problems 6–3
AP 6.6 [a] Summing the voltages around mesh 1 yields
4di1dt
+ 8d(i2 + ig)
dt+ 20(i1 − i2) + 5(i1 + ig) = 0
or
4di1dt
+ 25i1 + 8di2dt
− 20i2 = −(
5ig + 8digdt
)
Summing the voltages around mesh 2 yields
16d(i2 + ig)
dt+ 8
di1dt
+ 20(i2 − i1) + 780i2 = 0
or
8di1dt
− 20i1 + 16di2dt
+ 800i2 = −16digdt
[b] From the solutions given in part (b)
i1(0) = −0.4 − 11.6 + 12 = 0; i2(0) = −0.01 − 0.99 + 1 = 0
These values agree with zero initial energy in the circuit. At infinity,
i1(∞) = −0.4A; i2(∞) = −0.01A
When t = ∞ the circuit reduces to
·. . i1(∞) = −(
7.8
20+
7.8
780
)
= −0.4A; i2(∞) = − 7.8
780= −0.01A
From the solutions for i1 and i2 we have
di1dt
= 46.40e−4t − 60e−5t
di2dt
= 3.96e−4t − 5e−5t
Also,digdt
= 7.84e−4t
Thus
4di1dt
= 185.60e−4t − 240e−5t
25i1 = −10 − 290e−4t + 300e−5t
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6–4 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
8di2dt
= 31.68e−4t − 40e−5t
20i2 = −0.20 − 19.80e−4t + 20e−5t
5ig = 9.8 − 9.8e−4t
8digdt
= 62.72e−4t
Test:
185.60e−4t − 240e−5t − 10 − 290e−4t + 300e−5t + 31.68e−4t − 40e−5t
+0.20 + 19.80e−4t − 20e−5t ?= −[9.8 − 9.8e−4t + 62.72e−4t]
−9.8 + (300 − 240 − 40 − 20)e−5t
+(185.60 − 290 + 31.68 + 19.80)e−4t ?= −(9.8 + 52.92e−4t)
−9.8 + 0e−5t + (237.08 − 290)e−4t ?= −9.8 − 52.92e−4t
−9.8 − 52.92e−4t = −9.8 − 52.92e−4t (OK)
Also,
8di1dt
= 371.20e−4t − 480e−5t
20i1 = −8 − 232e−4t + 240e−5t
16di2dt
= 63.36e−4t − 80e−5t
800i2 = −8 − 792e−4t + 800e−5t
16digdt
= 125.44e−4t
Test:
371.20e−4t − 480e−5t + 8 + 232e−4t − 240e−5t + 63.36e−4t − 80e−5t
−8 − 792e−4t + 800e−5t ?= −125.44e−4t
(8 − 8) + (800 − 480 − 240 − 80)e−5t
+(371.20 + 232 + 63.36 − 792)e−4t ?= −125.44e−4t
(800 − 800)e−5t + (666.56 − 792)e−4t ?= −125.44e−4t
−125.44e−4t = −125.44e−4t (OK)
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Problems 6–5
Problems
P 6.1 [a] v = Ldi
dt
= (150 × 10−6)(25)[e−500t − 500te−500t] = 3.75e−500t(1 − 500t)mV
[b] i(5ms) = 25(0.005)(e−2.5) = 10.26mA
v(5ms) = 0.00375(e−2.5)(1 − 2.5) = −461.73µV
p(5ms) = vi = (10.26 × 10−3)(−461.73 × 10−6) = −4.74µW
[c] delivering 4.74µW
[d] i(5ms) = 10.26mA (from part [b])
w =1
2Li2 =
1
2(150 × 10−6)(0.01026)2 = 7.9nJ
[e] The energy is a maximum where the current is a maximum:
diLdt
= 0 when 1 − 500t = 0 or t = 2ms
imax = 25(0.002)e−1 = 18.39mA
wmax =1
2(150 × 10−6)(0.01839)2 = 25.38nJ
P 6.2 [a] i = 0 t < 0
i = 4t A 0 ≤ t ≤ 25ms
i = 0.2 − 4t A 25 ≤ t ≤ 50ms
i = 0 50ms < t
[b] v = Ldi
dt= 500 × 10−3(4) = 2V 0 ≤ t ≤ 25ms
v = 500 × 10−3(−4) = −2V 25 ≤ t ≤ 50ms
v = 0 t < 0
v = 2V 0 < t < 25ms
v = −2V 25 < t < 50ms
v = 0 50ms < t
p = vi
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6–6 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
p = 0 t < 0
p = (4t)(2) = 8t W 0 < t < 25ms
p = (0.2 − 4t)(−2) = 8t − 0.4W 25 < t < 50ms
p = 0 50ms < t
w = 0 t < 0
w =∫ t
0(8x) dx = 8
x2
2
∣
∣
∣
∣
t
0= 4t2 J 0 < t < 25ms
w =∫ t
0.025(8x − 0.4) dx + 2.5 × 10−3
= 4x2 − 0.4x∣
∣
∣
∣
t
0.025+2.5 × 10−3
= 4t2 − 0.4t + 10 × 10−3 J 25 < t < 50ms
w = 0 10ms < t
P 6.3 [a] i(0) = A1 + A2 = 0.12
di
dt= −500A1e
−500t − 2000A2e−2000t
v = −25A1e−500t − 100A2e
−2000t V
v(0) = −25A1 − 100A2 = 3
Solving, A1 = 0.2 and A2 = −0.08
Thus,
i = 200e−500t − 80e−2000t mA t ≥ 0
v = −5e−500t + 8e−2000t V, t ≥ 0
[b] i = 0 when 200e−500t = 80e−2000t
Therefore
e1500t = 0.4 so t = −610.86µs which is not possible!
v = 0 when 5e−500t = 8e−2000t
Therefore
e1500t = 1.6 so t = 313.34µs
Thus the power is zero at t = 313.34µs.
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Problems 6–7
P 6.4 [a] From Problem 6.3 we have
i = A1e−500t + A2e
−2000t A
v = −25A1e−500t − 100A2e
−2000t V
i(0) = A1 + A2 = 0.12
v(0) = −25A1 − 100A2 = −18
Solving, A1 = −0.08; A2 = 0.2Thus,
i = −80e−500t + 200e−2000t mA t ≥ 0
v = 2e−500t − 20e−2000t V t ≥ 0
[b] i = 0 when 80e−500t = 200e−2000t
·. . e1500t = 2.5 so t = 610.86µs
Thus,
i > 0 for 0 ≤ t < 610.86µs and i < 0 for 610.86µs < t < ∞
v = 0 when 2e−500t = 20e−2000t
·. . e1500t = 10 so t = 1535.06µs
Thus,
v < 0 for 0 ≤ t < 1535.06µs and v > 0 for 1535.06µs < t < ∞Therefore,
p < 0 for 0 ≤ t < 610.86µs and 1535.06µs < t < ∞
(inductor delivers energy)
p > 0 for 610.86µs < t < 1535.06µs (inductor stores energy)
[c] The energy stored at t = 0 is
w(0) =1
2L[i(0)]2 =
1
2(0.05)(0.12)2 = 360µJ
p = vi = −0.16e−1000t + 2e−2500t − 4e−4000t W
For t > 0:
w =∫
∞
0− 0.16e−1000t dt +
∫
∞
02e−2500t dt −
∫
∞
04e−4000t dt
=−0.16e−1000t
−1000
∣
∣
∣
∣
∞
0+
2e−2500t
−2500
∣
∣
∣
∣
∞
0−4e−4000t
−4000
∣
∣
∣
∣
∞
0
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6–8 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
= (−1.6 + 8 − 10) × 10−4
= −360µJ
Thus, the energy stored equals the energy extracted.
P 6.5 i = (B1 cos 200t + B2 sin 200t)e−50t
i(0) = B1 = 75mA
di
dt= (B1 cos 200t + B2 sin 200t)(−50e−50t) + e−50t(−200B1 sin 200t + 200B2 cos 200t)
= [(200B2 − 50B1) cos 200t − (200B1 + 50B2) sin 200t]e−50t
v = 0.2di
dt= [(40B2 − 10B1) cos 200t − (40B1 + 10B2) sin 200t]e−50t
v(0) = 4.25 = 40B2 − 10B1 = 40B2 − 0.75 ·. . B2 = 125mA
Thus,
i = (75 cos 200t + 125 sin 200t)e−50t mA, t ≥ 0
v = (4.25 cos 200t − 4.25 sin 200t)e−50t V, t ≥ 0
i(0.025) = −28.25mA; v(0.025) = 1.513V
p(0.025) = (−28.25)(1.513) = −42.7mW delivering
P 6.6 p = vi = 40t[e−10t − 10te−20t − e−20t]
W =∫
∞
0p dx =
∫
∞
040x[e−10x − 10xe−20x − e−20x] dx = 0.2 J
This is energy stored in the inductor at t = ∞.
P 6.7 [a] 0 ≤ t ≤ 50ms :
i =1
L
∫ t
0vs dx + i(0) =
106
750
∫ t
00.15dx + 0
= 200x
∣
∣
∣
∣
t
0= 200t A
i(0.05) = 200(0.05) = 10A
t ≥ 50ms : i =106
750
∫ t
50×10−3
(0) dx + 10 = 10A
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Problems 6–9
[b] i = 200t A, 0 ≤ t ≤ 50ms; i = 10A, t ≥ 50ms
P 6.8 0 ≤ t ≤ 100ms :
iL =103
50
∫ t
02e−100x dx + 0.1 = 40
e−100x
−100
∣
∣
∣
∣
t
0+ 0.1
= −0.4e−100t + 0.5A
iL(0.1) = −0.4e−10 + 0.5 = 0.5A
t ≥ 100ms :
iL =103
50
∫ t
0.1−2e−100(x−0.1) dx + 0.5 = −40
e−100(x−0.1)
−100
∣
∣
∣
∣
t
0.1+ 0.5
= 0.4e−100(t−0.1) + 0.1A
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6–10 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P 6.9 [a] 0 ≤ t ≤ 25ms :
v = 800t
i =1
10
∫ t
0800x dx + 0 = 80
x2
2
∣
∣
∣
∣
t
0
i = 40t2 A
25ms ≤ t ≤ 75ms :
v = 20
i(0.025) = 25mA
·. . i =1
10
∫ t
0.02520dx + 0.025
= 2x
∣
∣
∣
∣
t
0.025+ 0.025
= 2t − 0.025A
75ms ≤ t ≤ 125ms :
v = 80 − 800t V
i(0.075) = 2(0.075) − 0.025 = 0.125A
i =1
10
∫ t
0.075(80 − 800x) dx + 0.125
=
(
8x − 80x2
2
)
∣
∣
∣
∣
t
0.075+ 0.125
= 8t − 40t2 − 0.25A
125ms ≤ t ≤ 150ms :
v = 800t − 120
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Problems 6–11
i(0.125) = 8(0.125) − 40(0.125)2 − 0.25 = 0.125A
i =1
10
∫ t
0.125(800x − 120) dx + 0.125
=
(
80x2
2− 12x
)
∣
∣
∣
∣
t
0.125+ 0.125
= 40t2 − 12t + 1A
t ≥ 150ms :
v = 0
i(0.150) = 40(0.15)2 − 12(0.15) + 1 = 0.1A
i =1
10
∫ t
0.150dx + 0.1
= 0.1A
[b] v = 0 at t = 100ms and t = 150ms
i(0.1) = 8(0.1) − 40(0.1)2 − 0.25 = 0.15A
i(0.15) = 0.1A
[c]
P 6.10 [a] i =1
0.1
∫ t
020 cos 80x dx
= 200sin 80x
80
∣
∣
∣
∣
t
0
= 2.5 sin 80t A
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6–12 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] p = vi = (20 cos 80t)(2.5 sin 80t)
= 50 cos 80t sin 80t
p = 25 sin 160t W
w =1
2Li2
=1
2(0.1)(2.5 sin 80t)2
= 312.5 sin2 80t mJ
w = (156.25 − 156.25 cos 160t)mJ
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Problems 6–13
[c] Absorbing power: Delivering power:
0 ≤ t ≤ π/160 s π/160 ≤ t ≤ π/80 s
π/80 ≤ t ≤ 3π/160 s 3π/160 ≤ t ≤ π/40 s
P 6.11 [a] v = Ldi
dt
v = −25 × 10−3 d
dt[10 cos 400t + 5 sin 400t]e−200t
= −25 × 10−3(−200e−200t[10 cos 400t + 5 sin 400t]
+e−200t[−4000 sin 400t + 2000 cos 400t])
v = −25 × 10−3e−200t(−1000 sin 400t − 4000 sin 400t)
= −25 × 10−3e−200t(−5000 sin 400t)
= 125e−200t sin 400t V
dv
dt= 125(e−200t(400) cos 400t − 200e−200t sin 400t)
= 25,000e−200t(2 cos 400t − sin 400t) V/s
dv
dt= 0 when 2 cos 400t = sin 400t
·. . tan 400t = 2, 400t = 1.11; t = 2.77ms
[b] v(2.77 ms) = 125e−0.55 sin 1.11 = 64.27V
P 6.12 For 0 ≤ t ≤ 1.6 s:
iL =1
5
∫ t
03 × 10−3 dx + 0 = 0.6 × 10−3t
iL(1.6 s) = (0.6 × 10−3)(1.6) = 0.96mA
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6–14 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
Rm = (20)(1000) = 20kΩ
vm(1.6 s) = (0.96 × 10−3)(20 × 103) = 19.2V
P 6.13 [a] i = Cdv
dt= (5 × 10−6)[500t(−2500)e−2500t + 500e−2500t]
= 2.5 × 10−3e−2500t(1 − 2500t)A
[b] v(100µ) = 500(100 × 10−6)e−0.25 = 38.94mV
i(100µ) = (2.5 × 10−3)e−0.25(1 − 0.25) = 1.46mA
p(100µ) = vi = (38.94 × 10−3)(1.46 × 10−3) = 56.86µW
[c] p > 0, so the capacitor is absorbing power.
[d] v(100µ) = 38.94mV
w =1
2Cv2 =
1
2(5 × 10−6)(38.94 × 10−3)2 = 3.79nJ
[e] The energy is maximum when the voltage is maximum:
dv
dt= 0 when (1 − 2500t) = 0 or t = 0.4ms
vmax = 500(0.4 × 10−3)2e−1 = 73.58mV
pmax =1
2Cv2
max = 13.53nJ
P 6.14 [a] v = 0 t < 0
v = 10t A 0 ≤ t ≤ 2 s
v = 40 − 10t A 2 ≤ t ≤ 6 s
v = 10t − 80A 6 ≤ t ≤ 8 s
v = 0 8 s < t
[b] i = Cdv
dt
i = 0 t < 0
i = 2mA 0 < t < 2 s
i = −2mA 2 < t < 6 s
i = 2mA 6 < t < 8 s
i = 0 8 s < t
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Problems 6–15
p = vi
p = 0 t < 0
p = (10t)(0.002) = 0.02t W 0 < t < 2 s
p = (40 − 10t)(−0.002) = 0.02t − 0.08W 2 < t < 6 s
p = (10t − 80)(0.002) = 0.02t − 0.16W 6 < t < 8 s
p = 0 8 s < t
w =∫
p dx
w = 0 t < 0
w =∫ t
0(0.02x) dx = 0.01x2
∣
∣
∣
∣
t
0= 0.01t2 J 0 < t < 2 s
w =∫ t
2(0.02x − 0.08) dx + 0.04
= (0.01x2 − 0.08x)
∣
∣
∣
∣
t
2+0.04
= 0.01t2 − 0.08t + 0.16 J 2 < t < 6 s
w =∫ t
6(0.02x − 0.16) dx + 0.04
= (0.01x2 − 0.16x)
∣
∣
∣
∣
t
6+ 0.04
= 0.01t2 − 0.16t + 0.64 J 6 < t < 8 s
w = 0 8 s < t
[c]
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6–16 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
From the plot of power above, it is clear that power is being absorbed for0 < t < 2 s and for 4 s < t < 6 s, because p > 0. Likewise, power is beingdelivered for 2 s < t < 4 s and 6 s < t < 8 s, because p < 0.
P 6.15 [a] w(0) =1
2C [v(0)]2 =
1
2(5 × 10−6)(60)2 = 9mJ
[b] v = (A1 + A2t)e−1500t
v(0) = A1 = 60V
dv
dt= −1500e−1500t(A1 + A2t) + e−1500t(A2)
= (−1500A2t − 1500A1 + A2)e−1500t
dv(0)
dt= A2 − 1500A1
i = Cdv
dt, i(0) = C
dv(0)
dt
·. .dv(0)
dt=
i(0)
C=
100 × 10−3
5 × 10−6= 20 × 103
·. . 20 × 103 = A2 − 1500(60)
Thus, A2 = 20 × 103 + 90 × 103 = 110 × 103 V
s
[c] v = (60 + 110 × 103t)e−1500t
i = Cdv
dt= 5 × 10−6 d
dt(60 + 110 × 103t)e−1500t
i = (5 × 10−6)[110,000e−1500t − 1500(60 + 110, 000t)e−1500t]
= (0.1 − 825t)e−1500t A, t ≥ 0
P 6.16 iC = C(dv/dt)
0 < t < 2 s : iC = 100 × 10−9(15)t2 = 1.5 × 10−6t2 A
2 < t < 4 s : iC = 100 × 10−9(−15)(t − 4)2 = −1.5 × 10−6(t − 4)2 A
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Problems 6–17
P 6.17 [a] i = Cdv
dt= 0, t < 0
[b] i = Cdv
dt= 120 × 10−6 d
dt[30 + 5e−500t(6 cos 2000t + sin 2000t)]
= 120 × 10−6[5(−500)e−500t(6 cos 2000t + sin 2000t)
+5(2000)e−500t(−6 sin 2000t + cos 2000t)]
= −0.6e−500t[cos 2000t + 12.5 sin 2000t] A, t ≥ 0
[c] no, v(0−) = 60 Vv(0+) = 30 + 5(6) = 60 V
[d] yes, i(0−) = 0 Ai(0+) = −0.6 A
[e] v(∞) = 30V
w =1
2Cv2 =
1
2(120 × 10−6)(30)2 = 54mJ
P 6.18 [a] v(20µs) = 12.5 × 109(20 × 10−6)2 = 5V (end of first interval)
v(20µs) = 106(20 × 10−6) − (12.5)(400) × 10−3 − 10
= 5V (start of second interval)
v(40µs) = 106(40 × 10−6) − (12.5)(1600) × 10−3 − 10
= 10V (end of second interval)
[b] p(10µs) = 62.5 × 1012(10−5)3 = 62.5mW, v(10µs) = 1.25V,
i(10µs) = 50mA, p(10µs) = vi = (1.25)(50m) = 62.5mW (checks)
p(30µs) = 437.50mW, v(30µs) = 8.75V, i(30µs) = 0.05A
p(30µs) = vi = (8.75)(0.05) = 62.5mW (checks)
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6–18 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[c] w(10µs) = 15.625 × 1012(10 × 10−6)4 = 0.15625µJ
w = 0.5Cv2 = 0.5(0.2 × 10−6)(1.25)2 = 0.15625µJ
w(30µs) = 7.65625µJ
w(30µs) = 0.5(0.2 × 10−6)(8.75)2 = 7.65625µJ
P 6.19 [a] v =1
0.5 × 10−6
∫ 500×10−6
050 × 10−3e−2000t dt − 20
= 100 × 103 e−2000t
−2000
∣
∣
∣
∣
500×10−6
0−20
= 50(1 − e−1) − 20 = 11.61V
w = 12Cv2 = 1
2(0.5)(10−6)(11.61)2 = 33.7µJ
[b] v(∞) = 50 − 20 = 30V
w(∞) =1
2(0.5 × 10−6)(30)2 = 225µJ
P 6.20 [a] i =5
2 × 10−3t = 2500t 0 ≤ t ≤ 2ms
i =−10
4 × 10−3t + 10 = 10 − 2500t 2 ≤ t ≤ 6ms
i =10
4 × 10−3t− 20 = 2500t − 20 6 ≤ t ≤ 10ms
i =−5
2 × 10−3t + 30 = 30 − 2500t 10 ≤ t ≤ 12ms
q =∫ 0.002
02500t dt +
∫ 0.006
0.002(10 − 2500t) dt
=2500t2
2
∣
∣
∣
∣
0.002
0+
(
10t − 2500t2
2
)
∣
∣
∣
∣
0.006
0.002
= 0.005 − 0 + (0.06 − 0.045) − (0.02 − 0.005)
= 5mC
[b] v = 0.5 × 106∫ 0.002
02500x dx + 0.5 × 106
∫ 0.006
0.002(10 − 2500x) dx
+0.5 × 106∫ 0.01
0.006(2500x − 20) dx
= 0.5 × 106
[
2500x2
2
∣
∣
∣
∣
0.002
0+10x
∣
∣
∣
∣
0.006
0.002−2500x2
2
∣
∣
∣
∣
0.006
0.002+
2500x2
2
∣
∣
∣
∣
0.01
0.006−20x
∣
∣
∣
∣
0.01
0.006
]
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Problems 6–19
= 0.5 × 106[(0.005 − 0) + (0.06 − 0.02) − (0.045 − 0.005)
+ (0.125 − 0.045) − (0.2 − 0.12)]
= 2500V
v(10ms) = 2500V
[c] v(12ms) = v(10ms) + 0.5 × 106∫ 0.012
0.01(30 − 2500x) dx
= 2500 + 0.5 × 106
(
30x − 2500x2
2
)
∣
∣
∣
∣
0.012
0.01
= 2500 + 0.5 × 106(0.36 − 0.18 − 0.3 + 0.125)
= 2500 + 2500 = 5000V
w =1
2Cv2 =
1
2(2 × 10−6)(5000)2 = 25J
P 6.21 [a] 0 ≤ t ≤ 10µs
C = 0.1µF1
C= 10 × 106
v = 10 × 106∫ t
0− 0.05dx + 15
v = −50 × 104t + 15V 0 ≤ t ≤ 10µs
v(10µs) = −5 + 15 = 10V
[b] 10µs ≤ t ≤ 20µs
v = 10 × 106∫ t
10×10−6
0.1dx + 10 = 106t − 10 + 10
v = 106t V 10 ≤ t ≤ 20µs
v(20µs) = 106(20 × 10−6) = 20V
[c] 20µs ≤ t ≤ 40µs
v = 10 × 106∫ t
20×10−6
1.6dx + 20 = 1.6 × 106t − 32 + 20
v = 1.6 × 106t − 12V, 20µs ≤ t ≤ 40µs
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6–20 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[d] 40µs ≤ t < ∞
v(40µs) = 64 − 12 = 52V 40µs ≤ t < ∞
P 6.22 [a] 15‖30 = 10mH
10 + 10 = 20mH
20‖20 = 10mH
12‖24 = 8mH
10 + 8 = 18mH
18‖9 = 6mH
Lab = 6 + 8 = 14mH
[b] 12 + 18 = 30µH
30‖20 = 12µH
12 + 38 = 50µH
30‖75‖50 = 15µH
15 + 15 = 30µH
30‖60 = 20µH
Lab = 20 + 25 = 45µH
P 6.23 [a] Combine two 10 mH inductors in parallel to get a 5 mH equivalentinductor. Then combine this parallel pair in series with three 1 mHinductors:
10m‖10m + 1m + 1m + 1m = 8mH
[b] Combine two 10µH inductors in parallel to get a 5µH inductor. Thencombine this parallel pair in series with four more 10µH inductors:
10µ‖10µ + 10µ + 10µ + 10µ + 10µ = 45µH
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Problems 6–21
[c] Combine two 100µH inductors in parallel to get a 50µH inductor. Thencombine this parallel pair with a 100µH inductor and three 10µHinductors in series:
100µ‖100µ + 100µ + 10µ + 10µ + 10µ = 180µH
P 6.24 [a]
3.2di
dt= 64e−4t so
di
dt= 20e−4t
i(t) = 20∫ t
0e−4x dx − 5
= 20e−4x
−4
∣
∣
∣
∣
t
0−5
i(t) = −5e−4t A
[b] 4di1dt
= 64e−4t
i1(t) = 16∫ t
0e−4x dx − 10
= 16e−4x
−4
∣
∣
∣
∣
t
0−10
i1(t) = −4e−4t − 6A
[c] 16di2dt
= 64e−4t sodi2dt
= 4e−4t
i2(t) = 4∫ t
0e−4x dx + 5
= 4e−4x
−4
∣
∣
∣
∣
t
0+5
i2(t) = −e−4t + 6A
[d] p = −vi = (−64e−4t)(−5e−4t) = 320e−8t W
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6–22 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
w =∫
∞
0p dt =
∫
∞
0320e−8t dt
= 320e−8t
−8
∣
∣
∣
∣
∞
0
= 40J
[e] w =1
2(4)(−10)2 +
1
2(16)(5)2 = 400J
[f ] wtrapped = winitial − wdelivered = 400 − 40 = 360 J
[g] wtrapped =1
2(4)(−6)2 +
1
2(16)(6)2 = 360J checks
P 6.25 [a] io(0) = −i1(0) − i2(0) = 6 − 1 = 5A
[b]
io = −1
4
∫ t
02000e−100x dx + 5 = −500
e−100x
−100
∣
∣
∣
∣
∣
t
0
+ 5
= 5(e−100t − 1) + 5 = 5e−100t A, t ≥ 0
[c]
va = 3.2(−500e−100t) = −1600e−100t V
vc = va + vb = −1600e−100t + 2000e−100t
= 400e−100t V
i1 =1
1
∫ t
0400e−100x dx − 6
= −4e−100t + 4 − 6
i1 = −4e−100t − 2A t ≥ 0
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Problems 6–23
[d] i2 =1
4
∫ t
0400e−100x dx + 1
= −e−100t + 2A, t ≥ 0
[e] w(0) =1
2(1)(6)2 +
1
2(4)(1)2 +
1
2(3.2)(5)2 = 60J
[f ] wdel =1
2(4)(5)2 = 50J
[g] wtrapped = 60 − 50 = 10J
or wtrapped =1
2(1)(2)2 +
1
2(4)(2)2 + 10J (check)
P 6.26 vb = 2000e−100t V
io = 5e−100t A
p = 10,000e−200t W
w =∫ t
0104e−200x dx = 10,000
e−200x
−200
∣
∣
∣
∣
t
0= 50(1 − e−200t)W
wtotal = 50J
80%wtotal = 40J
Thus,
50 − 50e−200t = 40; e200t = 5; ·. . t = 8.05ms
P 6.27 [a]1
C1=
1
48+
1
24=
1
16; C1 = 16nF
C2 = 4 + 16 = 20nF
1
C3=
1
20+
1
30=
1
12; C3 = 12nF
C4 = 12 + 8 = 20nF
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6–24 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
1
C5=
1
20+
1
20+
1
10=
1
5; C5 = 5nF
Equivalent capacitance is 5 nF with an initial voltage drop of +15 V.
[b]1
36+
1
18+
1
12=
1
6·. . Ceq = 6µF 24 + 6 = 30µF
25 + 5 = 30µF
1
30+
1
30+
1
30=
3
30·. . Ceq = 10µF
Equivalent capacitance is 10µF with an initial voltage drop of +25 V.
P 6.28 [a] Combine a 470 pF capacitor and a 10 pF capacitor in parallel to get a 480pF capacitor:
(470 p) in parallel with (10 p) = 470 p + 10 p = 480 pF
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Problems 6–25
[b] Create a 1200 nF capacitor as follows:
(1µ) in parallel with (0.1µ) in parallel with (0.1µ)
= 1000n + 100n + 100n = 1200nF
Create a second 1200 nF capacitor using the same three resistors. Placethese two 1200 nF in series:
(1200n) in series with (1200n) =(1200n)(1200n)
1200n + 1200n= 600nF
[a] Combine two 220µF capacitors in series to get a 110µF capacitor. Thencombine the series pair in parallel with a 10µF capacitor to get 120µF:
[(220µ) in series with (220µ)] in parallel with (10µ)
=(220µ)(220µ)
220µ + 220µ+ 10µ = 120µF
P 6.29 From Figure 6.17(a) we have
v =1
C1
∫ t
0i dx + v1(0) +
1
C2
∫ t
0i dx + v2(0) + · · ·
v =[
1
C1+
1
C2+ · · ·
]∫ t
0i dx + v1(0) + v2(0) + · · ·
Therefore1
Ceq=[
1
C1+
1
C2+ · · ·
]
, veq(0) = v1(0) + v2(0) + · · ·
P 6.30 From Fig. 6.18(a)
i = C1dv
dt+ C2
dv
dt+ · · · = [C1 + C2 + · · ·]dv
dt
Therefore Ceq = C1 + C2 + · · ·. Because the capacitors are in parallel, theinitial voltage on every capacitor must be the same. This initial voltage wouldappear on Ceq.
P 6.31 [a]
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6–26 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
vo =106
1.6
∫ t
0800 × 10−6e−25x dx − 20
= 500e−25x
−25
∣
∣
∣
∣
t
0−20
= −20e−25t V, t ≥ 0
[b] v1 =106
2(800 × 10−6)
e−25x
−25
∣
∣
∣
∣
t
0+5
= −16e−25t + 21V, t ≥ 0
[c] v2 =106
8(800 × 10−6)
e−25x
−25
∣
∣
∣
∣
t
0−25
= −4e−25t − 21V, t ≥ 0
[d] p = −vi = −(−20e−25t)(800 × 10−6)e−25t
= 16 × 10−3e−50t
w =∫
∞
016 × 10−3e−50t dt
= 16 × 10−3 e−50t
−50
∣
∣
∣
∣
∞
0
= −0.32 × 10−3(0 − 1) = 320µJ
[e] w = 12(2 × 10−6)(5)2 + 1
2(8 × 10−6)(25)2
= 2525µJ
[f ] wtrapped = winitial − wdelivered = 2525 − 320 = 2205µJ
[g] wtrapped = 12(2 × 10−6)(21)2 + 1
2(8 × 10−6)(−21)2
= 2205µJ
P 6.321
Ce
=1
1+
1
5+
1
1.25=
10
5= 2
·. . C2 = 0.5µF
vb = 20 − 250 + 30 = −200V
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Problems 6–27
[a]
vb = −106
0.5
∫ t
0− 5 × 10−3e−50x dx − 200
= 10,000e−50x
−50
∣
∣
∣
∣
t
0−200
= −200e−50t V
[b] va = −106
0.5
∫ t
0− 5 × 10−3e−50x dx − 20
= 20(e−50t − 1) − 20
= 20e−50t − 40V
[c] vc =106
1.25
∫ t
0− 5 × 10−3e−50x dx − 30
= 80(e−50t − 1) − 30
= 80e−50t − 110V
[d] vd = 106∫ t
0− 5 × 10−3e−50x dx + 250
= 100(e−50t − 1) + 250
= 100e−50t + 150V
CHECK: vb = −vc − vd − va
= −200e−50t V (checks)
[e] i1 = 0.2 × 10−6 d
dt[100e−50t + 150]
= 0.2 × 10−6(−5000e−50t) = −e−50t mA
[f ] i2 = 0.8 × 10−6 d
dt[100e−50t + 150] = −4e−50t mA
CHECK: ib = i1 + i2 = −5e−50t mA (OK)
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6–28 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P 6.33 [a] w(0) = 12(0.2 × 10−6)(250)2 + 1
2(0.8 × 10−6)(250)2 + 1
2(5 × 10−6)(20)2
+ 12(1.25 × 10−6)(30)2
= 32,812.5µJ
[b] w(∞) = 12(5 × 10−6)(40)2 + 1
2(1.25 × 10−6)(110)2 + 1
2(0.2 × 10−6)(150)2
+ 12(0.8 × 10−6)(150)2
= 22,812.5µJ
[c] w =1
2(0.5 × 10−6)(200)2 = 10,000µJ
CHECK: 32,812.5 − 22,812.5 = 10,000µJ
[d] % delivered =10,000
32,812.5× 100 = 30.48%
[e] w =∫ t
0(−0.005e−50x)(−200e−50x) dx =
∫ t
0e−100x dx
= 10(1 − e−100t)mJ
·. . 10−2(1 − e−100t) = 7.5 × 10−3; e−100t = 0.25
Thus, t =ln 4
100= 13.86ms.
P 6.34 vc = − 1
10 × 10−6
(∫ t
00.2e−800x dx −
∫ t
00.04e−200x dx
)
+ 5
= 25(e−800t − 1) − 20(e−200t − 1) + 5
= 25e−800t − 20e−200t V
vL = 150 × 10−3 diodt
= 150 × 10−3(−160e−800t + 8e−200t)
= −24e−800t + 1.2e−200t V
vo = vc − vL
= (25e−800t − 20e−200t) − (−24e−800t + 1.2e−200t)
= 49e−800t − 21.2e−200t V, t > 0
P 6.35diodt
= (2)e−5000t[−1000 sin 1000t + 5000 cos 1000t]
+(−5000e−5000t)[cos 1000t + 5 sin 1000t]= e−5000t− 52,000 sin 1000tV
diodt
(0+) = (1)[sin(0)] = 0
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Problems 6–29
·. . 50 × 10−3 diodt
(0+) = 0 so v2(0+) = 0
v1(0+) = 25io(0
+) + v2(0+) = 25(2) + 0 = 50V
P 6.36 [a] Rearrange by organizing the equations by di1/dt, i1, di2/dt, i2 and transferthe ig terms to the right hand side of the equations. We get
4di1dt
+ 25i1 − 8di2dt
− 20i2 = 5ig − 8digdt
−8di1dt
− 20i1 + 16di2dt
+ 80i2 = 16digdt
[b] From the given solutions we have
di1dt
= −320e−5t + 272e−4t
di2dt
= 260e−5t − 204e−4t
Thus,
4di1dt
= −1280e−5t + 1088e−4t
25i1 = 100 + 1600e−5t − 1700e−4t
8di2dt
= 2080e−5t − 1632e−4t
20i2 = 20 − 1040e−5t + 1020e−4t
5ig = 80 − 80e−5t
8digdt
= 640e−5t
Thus,
−1280e−5t + 1088e−4t + 100 + 1600e−5t − 1700e−4t − 2080e−5t
+1632e−4t − 20 + 1040e−5t − 1020e−4t ?= 80 − 80e−5t − 640e−5t
80 + (1088 − 1700 + 1632 − 1020)e−4t
+(1600 − 1280 − 2080 + 1040)e−5t ?= 80 − 720e−5t
80 + (2720 − 2720)e−4t + (2640 − 3360)e−5t = 80 − 720e−5t (OK)
8di1dt
= −2560e−5t + 2176e−4t
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6–30 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
20i1 = 80 + 1280e−5t − 1360e−4t
16di2dt
= 4160e−5t − 3264e−4t
80i2 = 80 − 4160e−5t + 4080e−4t
16digdt
= 1280e−5t
2560e−5t − 2176e−4t − 80 − 1280e−5t + 1360e−4t + 4160e−5t − 3264e−4t
+80 − 4160e−5t + 4080e−4t ?= 1280e−5t
(−80 + 80) + (2560 − 1280 + 4160 − 4160)e−5t
+(1360 − 2176 − 3264 + 4080)e−4t ?= 1280e−5t
0 + 1280e−5t + 0e−4t = 1280e−5t (OK)
P 6.37 [a] Yes, using KVL around the lower right loop
vo = v20Ω + v60Ω = 20(i2 − i1) + 60i2
[b] vo = 20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t)+
60(1 − 52e−5t + 51e−4t)
= 20(−3 − 116e−5t + 119e−4t) + 60 − 3120e−5t + 3060e−4t
vo = −5440e−5t + 5440e−4t V
[c] vo = L2d
dt(ig − i2) + M
di1dt
= 16d
dt(15 + 36e−5t − 51e−4t) + 8
d
dt(4 + 64e−5t − 68e−4t)
= −2880e−5t + 3264e−4t − 2560e−5t + 2176e−4t
vo = −5440e−5t + 5440e−4t V
P 6.38 [a] vg = 5(ig − i1) + 20(i2 − i1) + 60i2
= 5(16 − 16e−5t − 4 − 64e−5t + 68e−4t)+
20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t)+
60(1 − 52e−5t + 51e−4t)
= 60 + 5780e−4t − 5840e−5t V
[b] vg(0) = 60 + 5780 − 5840 = 0V
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Problems 6–31
[c] pdev = vgig
= 960 + 92,480e−4t − 94,400e−5t − 92,480e−9t+
93,440e−10tW
[d] pdev(∞) = 960W
[e] i1(∞) = 4A; i2(∞) = 1A; ig(∞) = 16A;
p5Ω = (16 − 4)2(5) = 720W
p20Ω = 32(20) = 180W
p60Ω = 12(60) = 60W∑
pabs = 720 + 180 + 60 = 960W
·. .∑
pdev =∑
pabs = 960W
P 6.39 [a] 0.5digdt
+ 0.2di2dt
+ 10i2 = 0
0.2di2dt
+ 10i2 = −0.5digdt
[b] i2 = 625e−10t − 250e−50t mA
di2dt
= −6.25e−10t + 12.5e−50t A/s
ig = e−10t − 10A
digdt
= −10e−10t A/s
0.2di2dt
+ 10i2 = 5e−10t and − 0.5digdt
= 5e−10t
[c] v1 = 5digdt
+ 0.5di2dt
= 5(−10e−10t) + 0.5(−6.25e−10t + 12.5e−50t)
= −53.125e−10t + 6.25e−50t V, t > 0
[d] v1(0) = −53.125 + 6.25 = −46.875V; Also
v1(0) = 5digdt
(0) + 0.5di2dt
(0)
= 5(−10) + 0.5(−6.25 + 12.5) = −46.875V
Yes, the initial value of v1 is consistent with known circuit behavior.
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6–32 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P 6.40 [a] vab = L1di
dt+ L2
di
dt+ M
di
dt+ M
di
dt= (L1 + L2 + 2M)
di
dt
It follows that Lab = (L1 + L2 + 2M)
[b] vab = L1di
dt− M
di
dt+ L2
di
dt− M
di
dt= (L1 + L2 − 2M)
di
dt
Therefore Lab = (L1 + L2 − 2M)
P 6.41 [a] vab = L1d(i1 − i2)
dt+ M
di2dt
0 = L1d(i2 − i1)
dt− M
di2dt
+ Md(i1 − i2)
dt+ L2
di2dt
Collecting coefficients of [di1/dt] and [di2/dt], the two mesh-currentequations become
vab = L1di1dt
+ (M − L1)di2dt
and
0 = (M − L1)di1dt
+ (L1 + L2 − 2M)di2dt
Solving for [di1/dt] gives
di1dt
=L1 + L2 − 2M
L1L2 − M2vab
from which we have
vab =
(
L1L2 − M2
L1 + L2 − 2M
)(
di1dt
)
·. . Lab =L1L2 − M2
L1 + L2 − 2M
[b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses,therefore
Lab =L1L2 − M2
L1 + L2 + 2M
P 6.42 When the switch is opened the induced voltage is negative at the dottedterminal. Since the voltmeter kicks downscale, the induced voltage across thevoltmeter must be negative at its positive terminal. Therefore, the voltage isnegative at the positive terminal of the voltmeter.
Thus, the upper terminal of the unmarked coil has the same instantaneouspolarity as the dotted terminal. Therefore, place a dot on the upper terminalof the unmarked coil.
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Problems 6–33
P 6.43 [a] Dot terminal 1; the flux is up in coil 1-2, and down in coil 3-4. Assign thecurrent into terminal 4; the flux is down in coil 3-4. Therefore, dotterminal 4. Hence, 1 and 4 or 2 and 3.
[b] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4.Assign the current into terminal 4; the flux is right-to-left in coil 3-4.Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3.
[c] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4.Assign the current into terminal 4; the flux is right-to-left in coil 3-4.Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3.
[d] Dot terminal 1; the flux is down in coil 1-2, and down in coil 3-4. Assignthe current into terminal 4; the flux is down in coil 3-4. Therefore, dotterminal 4. Hence, 1 and 4 or 2 and 3.
P 6.44 [a]1
k2=(
1 +P11
P12
)(
1 +P22
P12
)
=(
1 +P11
P21
)(
1 +P22
P12
)
Therefore
k2 =P12P21
(P21 + P11)(P12 + P22)
Now note that
φ1 = φ11 + φ21 = P11N1i1 + P21N1i1 = N1i1(P11 + P21)
and similarly
φ2 = N2i2(P22 + P12)
It follows that
(P11 + P21) =φ1
N1i1
and
(P22 + P12) =
(
φ2
N2i2
)
Therefore
k2 =(φ12/N2i2)(φ21/N1i1)
(φ1/N1i1)(φ2/N2i2)=
φ12φ21
φ1φ2
or
k =
√
√
√
√
(
φ21
φ1
)(
φ12
φ2
)
[b] The fractions (φ21/φ1) and (φ12/φ2) are by definition less than 1.0,therefore k < 1.
P 6.45 [a] k =M√L1L2
=22.8√576
= 0.95
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6–34 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] Mmax =√
576 = 24mH
[c]L1
L2=
N21P1
N22P2
=(
N1
N2
)2
·. .(
N1
N2
)2
=60
9.6= 6.25
N1
N2=
√6.25 = 2.5
P 6.46 [a] L2 =
(
M2
k2L1
)
=(0.09)2
(0.75)2(0.288)= 50mH
N1
N2=
√
L1
L2=
√
288
50= 2.4
[b] P1 =L1
N21
=0.288
(1200)2= 0.2 × 10−6 Wb/A
P2 =L2
N22
=0.05
(500)2= 0.2 × 10−6 Wb/A
P 6.47 [a] W = (0.5)L1i21 + (0.5)L2i
22 + Mi1i2
M = 0.85√
(18)(32) = 20.4mH
W = [9(36) + 16(81) + 20.4(54)] = 2721.6mJ
[b] W = [324 + 1296 + 1101.6] = 2721.6mJ
[c] W = [324 + 1296 − 1101.6] = 518.4mJ
[d] W = [324 + 1296 − 1101.6] = 518.4mJ
P 6.48 [a] M = 1.0√
(18)(32) = 24mH, i1 = 6A
Therefore 16i22 + 144i2 + 324 = 0, i22 + 9i2 + 20.25 = 0
Therefore i2 = −(
9
2
)
±√
(
9
2
)2
− 20.25 = −4.5 ±√
0
Therefore i2 = −4.5A
[b] No, setting W equal to a negative value will make the quantity under thesquare root sign negative.
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Problems 6–35
P 6.49 [a] L1 = N21P1; P1 =
72 × 10−3
6.25 × 104= 1152 nWb/A
dφ11
dφ21=
P11
P21= 0.2; P21 = 2P11
·. . 1152 × 10−9 = P11 + P21 = 3P11
P11 = 192 nWb/A; P21 = 960 nWb/A
M = k√
L1L2 = (2/3)√
(0.072)(0.0405) = 36mH
N2 =M
N1P21=
36 × 10−3
(250)(960 × 10−9)= 150 turns
[b] P2 =L2
N22
=40.5 × 10−3
(150)2= 1800 nWb/A
[c] P11 = 192 nWb/A [see part (a)]
[d]φ22
φ12
=P22
P12
=P2 − P12
P12
=P2
P12
− 1
P21 = P21 = 960 nWb/A; P2 = 1800 nWb/A
φ22
φ12=
1800
960− 1 = 0.875
P 6.50 P1 =L1
N21
= 2 nWb/A; P2 =L2
N22
= 2 nWb/A; M = k√
L1L2 = 180µH
P12 = P21 =M
N1N2= 1.2 nWb/A
P11 = P1 − P21 = 0.8 nWb/A
P 6.51 When the touchscreen in the mutual-capacitance design is touched at the pointx, y, the touch capacitance Ct is present in series with the mutual capacitanceat the touch point, Cmxy. Remember that capacitances combine in series theway that resistances combine in parallel. The resulting mutual capacitance is
C ′
mxy =CmxyCt
Cmxy + Ct
P 6.52 [a] The self-capacitance and the touch capacitance are effectively connected inparallel. Therefore, the capacitance at the x-grid electrode closest to thetouch point with respect to ground is
Cx = Cp + Ct = 30pF + 15pF = 45pF.
The same capacitance exists at the y-grid electrode closest to the touchpoint with respect to ground.
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6–36 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] The mutual-capacitance and the touch capacitance are effectivelyconnected in series. Therefore, the mutual capacitance between thex-grid and y-grid electrodes closest to the touch point is
C ′
mxy =CmxyCt
Cmxy + Ct
=(30)(15)
30 + 15= 10pF.
[c] In the self-capacitance design, touching the screen increases thecapacitance being measured at the point of touch. For example, in part(a) the measured capacitance before the touch is 30 pF and after thetouch is 45 pF. In the mutual-capacitance design, touching the screendecreases the capacitance being measured at the point of touch. Forexample, in part (b) the measured capacitance before the touch is 30 pFand after the touch is 10 pF.
P 6.53 [a] The four touch points identified are the two actual touch points and twoghost touch points. Their coordinates, in inches from the upper leftcorner of the screen, are
(2.1, 4.3); (3.2, 2.5); (2.1, 2.5); and (3.2, 4.3)
These four coordinates identify a rectangle within the screen, shownbelow.
[b] The touch points identified at time t1 are those listed in part (a). Thetouch points recognized at time t2 are
(1.8, 4.9); (3.9, 1.8); (1.8, 1.8); and (3.9, 4.9)
The first two coordinates are the actual touch points and the last twocoordinates are the associated ghost points. Again, the four coordinatesidentify a rectangle at time t2, as shown here:
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Problems 6–37
Note that the rectangle at time t2 is larger than the rectangle at time t1,so the software would recognize the two fingers are moving toward theedges of the screen. This pinch gesture thus specifies a zoom-in for thescreen.
[c] The touch points identified at time t1 are those listed in part (a). Thetouch points recognized at time t2 are
(2.8, 3.9); (3.0, 2.8); (2.8, 2.8); and (3.0, 3.9)
The first two coordinates are the actual touch points and the last twocoordinates are the associated ghost points. Again, the four coordinatesidentify a rectangle at time t2, as shown here:
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6–38 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
Here, the rectangle at time t2 is smaller than the rectangle at time t1, sothe software would recognize the two fingers are moving toward themiddle of the screen. This pinch gesture thus specifies a zoom-out for thescreen.
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